Question

The integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}\mathrm{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given: $\frac{dy}{dx}+y=\frac{1+y}{x}$


$$\begin{gathered} \frac{dy}{dx}+y=\frac{1+y}{x} \\ \Rightarrow \frac{dy}{dx}+y=\frac{1}{x}+\frac{y}{x} \\ \Rightarrow \frac{dy}{dx}+y-\frac{y}{x}=\frac{1}{x} \\ \Rightarrow \frac{dy}{dx}+y\left(1-\frac{1}{x}\right)=\frac{1}{x} \\ \Rightarrow \frac{dy}{dx}+y\left(\frac{x-1}{x}\right)=\frac{1}{x} \\ \mathrm{Here},P=\left(\frac{x-1}{x}\right)\mathrm{and}Q=\frac{1}{x} \\ \mathrm{Integrating}\mathrm{factor}=e^{\int Pdx} \\ =e^{\int \left(\frac{x-1}{x}\right)dx} \\ =e^{\int \left(1-\frac{1}{x}\right)dx} \\ =e^{x-\log x} \\ =e^x{e}^{-\log x} \\ =e^x{e}^{\log x^{-1}} \\ =e^x{x}^{-1} \\ =\frac{e^x}{x} \end{gathered}$$


Hence, the integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}\mathrm{is}\underline{\frac{e^x}{x}}.$